We now find the matrix of representation of $T$ relative to these bases. This will be a $4 \times 4$ matrix whose $i$-th row is the column vector of $T(e_i)$ relative to the basis $C$. For example, we have
\[ T(e_1) = 2 – x^2 + x^3 = 2p_1 – p_3 + p_4 , \]
and so the first column of the matrix is the column vector
\[ [T(e_1)]_{C} = \begin{bmatrix} 2 \\ 0 \\ -1 \\ 1 \end{bmatrix} . \]

This reduced matrix has three non-zero rows, and so has rank 3. Thus the orginal linear transformation $T$ also has rank 3. The nullity can be computed using the equation
\[ \mathtt{rank}(T) + \mathtt{nullity}(T) = \dim (V).\]
Using the values $\mathtt{rank}(T) = 3$ and $\dim (V) = 4$, the nullity of $T$ must be 1.

Linear Transformation to 1-Dimensional Vector Space and Its Kernel
Let $n$ be a positive integer. Let $T:\R^n \to \R$ be a non-zero linear transformation.
Prove the followings.
(a) The nullity of $T$ is $n-1$. That is, the dimension of the nullspace of $T$ is $n-1$.
(b) Let $B=\{\mathbf{v}_1, \cdots, \mathbf{v}_{n-1}\}$ be a basis of the […]